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Joint Probability

Jointprobabilityis used in multi-stage experiments when we want to find out how likely it is that two (or more) events happen at the same time. An example is drawing three cards and getting all jacks P (JJJ). We can use a tree diagram to figure out this type ofprobability. This type ofprobabilitydepends on whether the experiment is done with or without replacement.

With replacement means that the object chosen on one stage is returned to the sample space before the next choice is made.Independentevents are considered with replacement since they do not affect each other, interfere with each other, or cause each other. For example, tossing a head on the first toss does not affect the outcome of flipping the coin a second time.

Theprobabilitythatindependentevents A and B happen simultaneously is found by using the multiplication rule, or the product of the individual probabilities.

Without replacement uses the same idea, only we consider the change in the sample space if the first choice is not replaced. These can be considered dependent events, since changing the sample space changes theprobability. We still use the multiplication rule, but the numerator and/or denominator decrease(s) for each of the stages.

Jointprobabilityis not the same as conditionalprobability, though the two concepts are often confused.Probabilitydefinitions can find their way into CFA exam questions. Naturally, there may also be questions that test the ability to calculate joint probabilities. Such computations require use of the multiplication rule, which states that the jointprobabilityof A and B is the product of the conditionalprobabilityof A given B, times theprobabilityof B. Inprobabilitynotation:

Multiplication rule: P (AB) = P(A | B) * P(B)

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Example:

Use the joint distribution for the pair of randomvariablesto calculate the expected value and variance of X and the expected value and variance of Y.

The jointprobabilityp(x,y)= P(X=x,Y=y)p(0,1)= .1 p(1,1)= .2

P (0, 0) =.4 p (1, 0) =.3

Is used to calculate p(z) , where z is a value of the rv Z=X+Y:P9Z=0)= p(0,0)= .4

P (Z=1) = p (0, 1) + p (1, 0) = .1+3= .4

P (Z=2) = p(1,1)= .2

E(X+Y) = E(Z)

= 0(.4)+ 1(.4)+ 2(.2)= .8

V(X+Y)= V(Z)

= (0-,.8)^2(.4)+1(1-,.8)^2(.4)+ (2-,.8)^2(.2)= .56

The randomvariablesX and Y are notindependent, because P(X=x and Y=y) does not equal P(X=x)P(Y=y). If X and Y had beenindependent, we would have found V(X+Y)=V(X)+V(Y). ForindependentrandomvariablesX and Y, Cov(X,Y)=0. In general, the covariance of X and Y is not zero. Above it is .05. Get latest updates on the related subject inStatistichomework helpandassignment helpattranstutors.com.

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